Quantum Harmonic Oscillator: Second Excited State Analysis
This solution details the calculation for the energy ($E$) and degeneracy ($d$) of the second excited state for a three-dimensional (3D), isotropic quantum harmonic oscillator.
Energy Calculation for 3D Oscillator
The energy levels of a 3D isotropic quantum harmonic oscillator are given by the formula:
$E_n = \left(n + \frac{3}{2}\right)\hbar\omega$
where $n$ is the principal quantum number ($n = n_1 + n_2 + n_3$), and $n_1, n_2, n_3$ are the quantum numbers for each dimension.
- The ground state corresponds to $n=0$.
- The first excited state corresponds to $n=1$.
- The second excited state corresponds to $n=2$.
Substituting $n=2$ into the energy formula:
$E_2 = \left(2 + \frac{3}{2}\right)\hbar\omega = \left(\frac{4}{2} + \frac{3}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega$
Thus, the energy of the second excited state is $E = \frac{7}{2}\hbar\omega$.
Degeneracy Calculation for 3D Oscillator
The degeneracy ($d_n$) for the $n$-th energy level of a 3D isotropic quantum harmonic oscillator is calculated using the formula:
$d_n = \frac{(n+D-1)!}{n!(D-1)!}$
For a 3D oscillator, $D=3$. We need the degeneracy for the second excited state, where $n=2$.
Substituting $n=2$ and $D=3$ into the degeneracy formula:
$d_2 = \frac{(2+3-1)!}{2!(3-1)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$
Therefore, the degeneracy of the second excited state is $d = 6$.
Conclusion
The energy and degeneracy for the second excited state ($n=2$) of a 3D isotropic quantum harmonic oscillator are:
$E = \frac{7}{2}\hbar\omega$ and $d = 6$.
This matches the first option.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ isA particle is subjected to a potential
$V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$
Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?- The wavefunction for a particle is given by the form $e^{-(iax+\beta)}$, where $a$ and $\beta$ are real constants. In which one of the following potentials $V(x)$, the particle is moving?
A particle of mass $m$ is moving in the potential
$V(x) = \begin{cases} V_0 + \frac{1}{2}m\omega_0^2x^2, & x > 0, \\ \infty, & x \le 0, \end{cases}$
Figures P, Q, R and S show different combinations of the values of $\omega_0$ and $V_0$.
$E_j^{(P)}, E_j^{(Q)}, E_j^{(R)}$ and $E_j^{(S)}$ with $j = 0, 1, 2, ...$, are the eigen-energies of the $j$-th level for the potentials shown in figures P, Q, R and S, respectively. Which of the statement is/are true?- Young's double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer) ?
